Could you prove me some hints to prove the following theorem?
If $A$ is a non-singular matrix, and $B$ is a matrix such that: $$ \|B-A\|<\frac{1}{\|A^{-1}\|}, $$ then $B$ is non-singular and that: $$ B^{-1}=A^{-1}\Sigma_{j=0}^{\infty}(I-BA^{-1})^{j} $$ Here, $\|\cdot\|$ denotes the matrix norm: $$ \|A\|=\sup\frac{\|Ax\|}{\|x\|} $$ There are two parts in this theorem: the first is to prove that $B$ is non-singular, and the second is to prove the formula for the inverse of $B$. I am stuck at both.
Note that from $\|B-A\|<\frac{1}{\|A^{-1}\|}$, we know
$$ \|I-BA^{-1}\|=\|(I-BA^{-1})AA^{-1}\|=\|(A-B)A^{-1}\|\leq\|(A-B)\|\|A^{-1}\| <1$$ So the geometric series of real numbers $$\Sigma_{j=0}^{\infty}\|(I-BA^{-1})\|^{j}$$ is converging, hence the corresponding matrix series is converging as well: $$M:=\Sigma_{j=0}^{\infty}(I-BA^{-1})^{j}$$
Furthermore let us do a direct computation showing $(I-BA^{-1})M = M-I$: $$(I-BA^{-1})M = (I-BA^{-1})\Sigma_{j=0}^{\infty}(I-BA^{-1})^{j} \\=\Sigma_{j=1}^{\infty}(I-BA^{-1})^{j}=\\ (I-BA^{-1})+ (I-BA^{-1})^2 + (I-BA^{-1})^3 +... =\\ (I + (I-BA^{-1})+ (I-BA^{-1})^2 + (I-BA^{-1})^3 +... ) -I =\\ \left(\Sigma_{j=0}^{\infty}(I-BA^{-1})^{j}\right)-I \\ =M-I$$
Thus $BA^{-1}M=I$, which shows that $A^{-1}M$ is the inverse of $B$.